In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch `(S_1)`. Also when `(S_1)` is opend and `(S_2)` is closed the capacitor is connected in series with inductor (L).

At the start, the capicitor was uncharged. when switch `(S_1)` is closed and `(S_2)` is kept open, the time constant of this circuit is `tau`. which of the following is correct

Options

A. after time interval `tau`, change on the capacitor is `CV//2`
B. after time interval `2tau`, charge on the capacitor is `CV(1-e^(-2))`
C. the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.
D. after time interval `2tau`, charge on the capacitor is `CV(1-e^(-1))`

Answer

B

Explanation

`Q=Q_(0)(1-e^(-t//tau))=CV(1-e^(-2))`